Question

You are about to start a cell biology experiment that requires a number of different chemical reagents. You have an unlimited supply of water and the following materials available in the lab to make your solutions; 500g dry glucose 1500 mL of 40% TRIS-buffer solution 150 mL of 0.3 molar ampicillin 2L of 35% SDS Using precise numerical values (to one decimal place, do NOT include units), describe how you will prepare 150 mL of 8% TRIS-buffer.

Answer 1

The answer is "212.5 ml"

Step-by-step explanation:

Please find the complete question in the attached file.

In 2 ml of
`35\%` SDS it makes 120 ml of
`7\%` SDS.

To mix x ml of
`35\%` SDS

`\to V_1S_1 = V_2S_2\\\\\to V_1 = x \ ml\\\\\to V_2 = 120\ ml\\\\\to S_1 = 35\% \\\\ \to S_2 = 7\%\\\\\to V_1 = (V_2S_2)/(S_1)\\\\`

`= ((120 * 7))/(35)\\\\=24 \ ml\\\\`

added water
`= 120-24 = 96\ ml`

So we have to add 24ml from
`35\%` SDS and add 96ml water to male 120ml
`7\%` SDS.

1500ml of 40% Tris Buffer solution

Let us assume we have to mix y ml of 40% Tris buffer solution

`\to V_1S_1 = V_2S_2\\\\\to V_1 = y \ ml\\\\\to V_2 = 250\ ml\\\\\to S_1 = 40\% \\\\ \to S_2 = 6\%\\\\\to V_1 = (V_2S_2)/(S_1)\\\\`

`= ((250 * 6))/(40)\\\\= 37.5\ ml`

added water
`= 250-37.5 = 212.5\ ml`

So we have to add 37.5 ml from 40% Tris Buffer and add 212.5 ml water to male 250 ml 6% Tris Buffer.

Answer 2

• We can prepare 150 mL of 8% TRIS-buffer by adding
  `30 ml` from
  `40\%` Tris Buffer and add
  `120 ml` water

Step-by-step explanation:

1500ml of 40% Tris Buffer solution

Let us assume we have to mix y ml of 40% Tris buffer solution

We know,

`V_1S_1 = V_2S_2`

where,

`V_1 = y ml\\\\V_2 = 150 ml\\\\S_1 = 40\% and S_2 = 8\%`

So,

`V1 = (V_2S_2)/(S_1)\\\\V_1 = (150*8)/(40)\\\\V_1 = 30 ml`

So water we have to add

`= 150-30\\\\= 120 ml`

So we have to add 30 ml from 40% Tris Buffer and add 120 ml water to male 250 ml 6% Tris Buffer.

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